Question

If the activity of ${}^{108}$$Ag$ is $3$ micro curie, the number of atoms present in it are ( $\lambda =$ 0.005 sec${}^{-1}$)

• A. 2.2 x 10${}^5$
• B. 2.2 x 10${}^6$
• C. 2.2 x 10${}^4$
• D. 2.2 x 10${}^7$

Answer 1

Answer: (D)

$\lambda =0.005se{c}^{-1}$
$1$ Micro curie $=3.7\times {10}^{4}dps$
Assuming all $Ag$ atoms to be radioactive, and each showing
Let no of atoms present $=1dpsno$
Given:
$\left(\frac{dN}{dt}\right)activity=\lambda N_0\Rightarrow 3.7\times 3\times {10}^{4}dps=0.005\times N_0$
$\Rightarrow N_0=2.2\times {10}^{7}atoms$